The negation would be "there exist a finite number of integers whose squareroot is irrational".

Just saying "there exist an infinite number of integers whose squareroot is rational" won't help because that is true! There exist an infinite number of integers whose squareroot is rational and an infinite number of integers whose squareroot is irrational.

But I don't know if a proof by contradiction is the best way to go. It would be sufficient to show that the squareroot of any prime is irrational.

== It follows that p*b^2 = a^2, and using the decomposition of naturals in product of primes and the definition of prime, deduce p divides a^2 and thus a ==> a = p^n*x, with (p,x) = 1 (i.e., n is the maximal power of p that divides a). But then

p = a^2/b^2 = p^(2n)*x^2/b^2 ==> b^2 = p^(2n-1)*x^2

== Deduce now that p divides both a, b contradicting that a/b is reduced.

As you can see, the above is basically the same argument used to prove that Sqrt(2) is not rational.

sqrt(p) is irrational for every prime p (because the only possible rational roots of polynomail x^2 - p are 1,-1,p,-p by rational roots theorem, thus sqrt(p) which is root of that polynomial must be irrational) ... and since there are infinite primes p, there are infinite sqrt(p) irrational numbers